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Math Help - Hemisphere angle of tilt where centre of mass near the edge mechanics help required

  1. #1
    Member ssadi's Avatar
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    Hemisphere angle of tilt where centre of mass near the edge mechanics help required

    The hemisphere of radius a has centre of mass at 3/2a, 1/6 a as shown in the diagram. How do I find the angle between PC and horizontal when the hemisphere is in equilibrium with its curved surface on ground?
    Attached Thumbnails Attached Thumbnails Hemisphere angle of tilt where centre of mass near the edge mechanics help required-untitled.png  
    Last edited by ssadi; May 23rd 2009 at 07:39 PM. Reason: typo
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  2. #2
    MHF Contributor chisigma's Avatar
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    The equilibrium condition is equivalent to minimum energy condition and that means that the ordinate of center of mass must be the lowest possibile, so that the angle between PC and horizontal [in radians] must be...

     \theta = tan^{-1} \frac {\frac{a}{2}}{\frac{a}{6}}= tan^{-1} 3 = 1,24904577...

    Kind regards

    \chi \sigma
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  3. #3
    Member ssadi's Avatar
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    Quote Originally Posted by chisigma View Post
    The equilibrium condition is equivalent to minimum energy condition and that means that the ordinate of center of mass must be the lowest possibile, so that the angle between PC and horizontal [in radians] must be...

     \theta = tan^{-1} \frac {\frac{a}{2}}{\frac{a}{6}}= tan^{-1} 3 = 1,24904577...

    Kind regards

    \chi \sigma
    can you please draw a picture of this? I am having some trouble imagining it.
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  4. #4
    MHF Contributor chisigma's Avatar
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    The equilibrium condition is also the 'minimum energy' condition and it is realized when the center of mass is at the minimum possible quote. The value of \theta I have indicated minimizes the quote of the center of mass, that means that for any other value of \theta the 'potential energy' is greater...



    Kind regards

    \chi \sigma
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